Q: Calculate the shortest and longest wavelengths of Balmer series of hydrogen atom. Given \$R=1.097\times 10^7m^{-1}\$.

Wavelength of photon emitted due to transition in H-atom \$\dfrac{1}{\lambda} = R \bigg( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \bigg)\$Shortest wavelength is emitted in Balmer series if the transition of electron takes place from \$n_2 = \infty\$ to \$n_1 = 2\$.\$\therefore\$ Shortest wavelength in Balmer series \$\dfrac{1}{\lambda_s} = R \bigg( \dfrac{1}{2^2} - \dfrac{1}{\infty} \bigg)\$Or \$\lambda_s = \dfrac{4}{R}\$\$\implies\$ \$\lambda_s = \dfrac{4}{1.097 \times 10^{7}} = 364.6 nm\$Longest wavelength is emitted in Balmer series if the transition of electron takes place from \$n_2 = 3\$ to \$n_1 = 2\$.\$\therefore\$ Longest wavelength in Balmer series \$\dfrac{1}{\lambda_L} = R \bigg( \dfrac{1}{2^2} - \dfrac{1}{3^2} \bigg)\$Or \$\lambda_s = \dfrac{36}{5R}\$\$\implies\$ \$\lambda_s = \dfrac{36}{5\times 1.097 \times 10^{7}} = 656.3 nm\$

Q: By assuming Bohr's postulates derive an expression for radius of \$n^{th}\$ orbit of electron, revolving round the nucleus of hydrogen atom.

Let the electron of mass \$m\$ revolves around a nucleus (H-atom) in an orbit of radius \$r_n\$ with linear velocity \$v_n\$.As the electron revolves in an stationary orbit, thus centrifugal force acting on the electron is balanced by the coulombic force i.e. \$F_{centrifugal} = F_{coulomb}\$\$\therefore\$ \$\dfrac{mv_n^2}{r_n} = \dfrac{ke^2}{r_n^2}\$ where \$k = \dfrac{1}{4\pi \epsilon_o} = 9\times 10^9\$\$\implies\$ \$r_n = \dfrac{ke^2}{mv_n^2}\$Also we use \$mv_nr_n = \dfrac{nh}{2\pi}\$Eliminating \$v_n\$ from both equations, we get \$r_n = \dfrac{ke^2}{m}.\dfrac{4\pi^2m^2 r_n^2}{n^2 h^2}\$\$\implies\$ \$r_n = \dfrac{h^2 n^2}{4\pi^2 m ke^2} \$Putting \$m = 9.1\times 10^{-31} kg\$, \$h = 6.626 \times 10^{-34} Js\$ and \$e = 1.6\times 10^{-19} C\$We get radius of \$n^{th}\$ Bohr orbit \$r_n = 0.529\$ \$ n^2\$ \$A^o\$

Q: The ground state energy of hydrogen atom is \$-13.6\ eV\$. What are the kinetic and potential energies of the electron in this state?

Ground state energy of hydrogen atom is \$E=-13.67eV\$The kinetic energy of the electron is given as:\$KE=-E\$\$=+13.67eV\$The potential energy of the electron is given as:\$ PE= - 2\times KE\$\$= - 27.2eV\$

Question 1

Write Bohr's postulates for the hydrogen atom model.

Accepted Answer

First Postulate: Electron revolves round the nucleus in discrete circular orbits called stationary orbits without emission of radiant energy. These orbits are called stable orbits or non-radiating orbits.Second Postulate: Electrons revolve around the nucleus only in orbits in which their angular momentum is an integral multiple of $h/2\pi$.Third Postulate: When an electron makes a transition from one of its non-radiating orbits to another of lower energy, a photon is emitted having energy equal to the energy difference between the two states. The frequency of the emitted photon is then given by, $v=\dfrac{{E}_{i}-{E}_{f}}{h}$

Question 2

Calculate the shortest and longest wavelengths of Balmer series of hydrogen atom. Given $R=1.097\times 10^7m^{-1}$.

Accepted Answer

Wavelength of photon emitted due to transition in H-atom $\dfrac{1}{\lambda} = R \bigg( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \bigg)$Shortest wavelength is emitted in Balmer series if the transition of electron takes place from $n_2 = \infty$ to $n_1 = 2$.$\therefore$ Shortest wavelength in Balmer series $\dfrac{1}{\lambda_s} = R \bigg( \dfrac{1}{2^2} - \dfrac{1}{\infty} \bigg)$Or $\lambda_s = \dfrac{4}{R}$$\implies$ $\lambda_s = \dfrac{4}{1.097 \times 10^{7}} = 364.6 nm$Longest wavelength is emitted in Balmer series if the transition of electron takes place from $n_2 = 3$ to $n_1 = 2$.$\therefore$ Longest wavelength in Balmer series $\dfrac{1}{\lambda_L} = R \bigg( \dfrac{1}{2^2} - \dfrac{1}{3^2} \bigg)$Or $\lambda_s = \dfrac{36}{5R}$$\implies$ $\lambda_s = \dfrac{36}{5\times 1.097 \times 10^{7}} = 656.3 nm$

Question 3

By assuming Bohr's postulates derive an expression for radius of $n^{th}$ orbit of electron, revolving round the nucleus of hydrogen atom.

Accepted Answer

Let the electron of mass $m$ revolves around a nucleus (H-atom) in an orbit of radius $r_n$ with linear velocity $v_n$.As the electron revolves in an stationary orbit, thus centrifugal force acting on the electron is balanced by the coulombic force i.e. $F_{centrifugal} = F_{coulomb}$$\therefore$ $\dfrac{mv_n^2}{r_n} = \dfrac{ke^2}{r_n^2}$ where $k = \dfrac{1}{4\pi \epsilon_o} = 9\times 10^9$$\implies$ $r_n = \dfrac{ke^2}{mv_n^2}$Also we use $mv_nr_n = \dfrac{nh}{2\pi}$Eliminating $v_n$ from both equations, we get  $r_n = \dfrac{ke^2}{m}.\dfrac{4\pi^2m^2 r_n^2}{n^2 h^2}$$\implies$ $r_n = \dfrac{h^2 n^2}{4\pi^2 m ke^2} $Putting $m = 9.1\times 10^{-31} kg$, $h = 6.626 \times 10^{-34} Js$ and $e = 1.6\times 10^{-19} C$We get radius of $n^{th}$ Bohr orbit  $r_n = 0.529$ $ n^2$ $A^o$

Question 4

Write two limitations of Bohr's model.

Accepted Answer

The Limitations of Bohr's model are,1) The Bohr atomic model theory made correct predictions for smaller sized atoms like hydrogen, but poor spectral predictions are obtained when larger atoms are considered.2) It failed to explain the Zeeman effect when the spectral line is split into several components in the presence of a magnetic field.

Question 5

The ground state energy of hydrogen atom is $-13.6\ eV$. What are the kinetic and potential energies of the electron in this state?

Accepted Answer

Ground state energy of hydrogen atom is $E=-13.67eV$The kinetic energy of the electron is given as:$KE=-E$$=+13.67eV$The potential energy of the electron is given as:$ PE= - 2\times KE$$= - 27.2eV$

Question 6

What is meant by 'Quantization of charge'?

Accepted Answer

Quantization of charge implies that charge can assume only certain discrete values. That is to say the observed value of electric charge (q) of a particle will be integral multiples of (e) $1.6\times10^{-19}$ coulombs. i.e.$q=ne$ where $n=0,1,2,....$ (both positive and negative integers)The charge cannot assume any value between the integers.

Question 7

Give the main postulates of Bohr's model of an atom.

Accepted Answer

Main postulates of Bohr's model of an atom are:1. In an atom, the electrons revolve around the nucleus in certain definite circular paths called orbits, or shells.2. Each shell or orbit corresponds to a definite energy. Therefore, these circular orbits are also known as energy levels or energy shells.3. The orbits or energy levels are characterized by an integer not, where, n can have values 1, 2, 3, 4&#8230;&#8230;. The integer not (= 1, 2, 3&#8230;) is called the quantum number of respective orbit. The orbits are numbered as 1, 2, 3, 4&#8230;&#8230;&#8230;&#8230; etc., starting from the nucleus side. Thus, the orbit for which n=1 is the lowest energy level.The orbits corresponding to n = 1,2,3,4&#8230;..etc., are also designated as K,L,M,N&#8230;&#8230;&#8230;.etc., shells. When the electron is in the lowest energy level, it is said to be in the ground state.Since, electronics can be present only in these orbits, hence, these electrons can only have energies corresponding to these energy levels, i.e., electrons in an atom can have only certain permissible energies .4. The electrons present in an atom can move from a lower energy level (Elower) to a level of higher energy (Ehigher) by absorbing the appropriate energy. Similarly, an electron can jump from a higher energy level (Ehigher) to a lower energy level (Elower) by losing the appropriate energy.The energy absorbed or lost is equal to the difference between the energies of the two energy levels, i.e.,&#916;E= E (higher) - E (lower)

Question 8

(a) Using Bohr's postulates, obtain the expression for total energy of the electron in the ${n}^{th}$ orbit of hydrogen atom.(b) What is the significance of negative sign in the expression for the energy?(c) Draw the energy level diagram showing how the line spectra corresponding to Paschen series occur due to transition between energy levels.

Accepted Answer

(a) According to Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge$+e$. For an electron moving with a uniform speed in a circular orbit on a given radius, the centripetal force is provided by the coulomb force of attraction between the electron and the nucleus.Therefore,$\frac{mv^2}{r}=\frac{1(Ze)(e)}{4\pi \epsilon_0 r^2}$$\implies$ $mv^2=\frac{1}{4\pi \epsilon_0 } \frac{Ze^2}{r}$     ....(1)So,Kinetic Energy, $ K.E=\frac12 mv^2$$K.E=\frac{1}{4\pi \epsilon_0} \frac{Ze^2}{r}$Potential Energy is given by, $P. E=\frac{1}{4\pi \epsilon_0} \frac{(Ze)(-e)}{r}$Therefore total energy is given by, $E=K.E+P.E$$E=\frac{1}{4 \pi \epsilon_0}\frac{Ze^2}{2r}+(-\frac{1}{4\pi \epsilon_0} \frac{Z e^2}{r})$$E=-\frac{1}{4 \pi \epsilon_0} \frac{Ze^2}{2r}$For, $n^{th}$ orbit, $E$ can be written as $E_n$,$E_n=-\frac{1}{4\pi \epsilon_0} \frac{Ze^2}{2r_n}$            ......(2)Now, using Bohr's postulate for quantization of angular momentum, we have$mvr=\frac{nh}{2 \pi}$$\implies$ $v= \frac{nh}{2 \pi mr}$Putting the values of $v$ in equation (1), we get$\frac{m}{r}$ $(\frac{nh}{2 \pi mr})^2=\frac{1}{4 \pi \epsilon_0} \frac{Ze^2}{r^2}$$=-\frac{mZ^2 e^4}{8 \epsilon_0 h^2 n^2}$$\implies$ $\frac{\epsilon_0 h^2 n^2}{\pi m Z e^2}$Now putting a]value of $r_n$ in equation (2), we get$E_n=-\frac{1}{4 \pi \epsilon_0} \frac{Ze^2}{2(\frac{\epsilon_0 h^2 n^2}{ \pi m})}$$=-\frac{mZ^2e^4}{8 \epsilon_0 h^2 n^2}$$\implies$ $E_n=-\frac{Z^2 Rhc}{n^2}$, where $R=\frac{m e^4}{8 \epsilon_0^2 c h^3}$where R is the Rydberg constant.(b)  Negative sign shows that electron remain bound with the nucleus.(c)  If electron jumps from $n_i=4,5,6 $ to $n_f=3$, the energy of the line spectra$\triangle E = \frac{me^4}{8 \epsilon_0^2 h^2} (\frac{1}{n_f^2}-\frac{1}{n_i^2})$

Question 9

Name spectral series of hydrogen which lies in the ultraviolet region of electromagnetic spectrum.

Accepted Answer

Lyman series of hydrogen atom lies in the ultraviolet region, Balmer series lies in visible region, Paschen series lies in near infrared region whereas Bracket, Pfund as well as Humphrey series lie in far infrared region of electromagnetic spectrum.

Question 10

Explain the significance of quantum number?

Accepted Answer

Quantum numbers are important because they can be used to determine the electron configuration of an atom and the probable location of the atom's electrons. Quantum numbers are also used to determine other characteristics of atoms, such as ionization energy and the atomic radius.